Problem: $h(r)=r^2+11r-26$ 1) What are the zeros of the function? Write the smaller $r$ first, and the larger $r$ second. $\text{smaller }r=$
Answer: To find the zeros of the function, we need to solve the equation $h(r)=0$. We can do that by factoring $h(r)$. $\begin{aligned} r^2+11r-26&=0 \\\\ (r-2)(r+13)&=0 \\\\ r-2=0&\text{ or }r+13=0 \\\\ r={2}&\text{ or }r={-13} \end{aligned}$ There are many ways to find the vertex. We will do it by using the fact that the $r$ -coordinate of the vertex is exactly between the two zeros. $\begin{aligned} \text{vertex's }r\text{-coordinate}&=\dfrac{({2})+({-13})}{2} \\\\ &=-{\dfrac{11}{2}} \end{aligned}$ Now we can find the vertex's $y$ -coordinate by evaluating $h\left({-\dfrac{11}{2}}\right)$ : $\begin{aligned} h\left({-\dfrac{11}{2}}\right)&=\left({-\dfrac{11}{2}}\right)^2+11\left(-{\dfrac{11}{2}}\right)-26 \\\\ &=\dfrac{121}{4}-\dfrac{121}{2}-26 \\\\ &=-\dfrac{225}{4} \end{aligned}$ In conclusion, $\begin{aligned} \text{smaller }r&=-13 \\\\ \text{larger }r&=2 \end{aligned}$ The vertex of the parabola is at $\left(-\dfrac{11}{2},-\dfrac{225}{4}\right)$